Homework 1

Homework # 1, Quantum Field Theory I: 7640
Due Monday, September 22
Problem 1. Quantization of EM field [15 points]
(a) Following discussion in the Lecture, we know that the vector potential A(r, t)
in the vacuum (no charges) obeys wave equation (in the Coulomb gauge) ∇2 A −
1 ∂2A
c2 ∂t2
= 0.
P P
k
α=1,2
√
Solution of this gives us the following mode decomposition Ak,α (r, t) =
1
20 ωk V
ak,α ek,α ei(k·r−ωk t) + c.c. , where ωk = ck and ak,α is the complex am-
plitude.
Starting from the energy of the EM field He.m. =
R
dr 20 E2 + c2 B2 and the flux of the
energy (Poynting vector) Se.m. = dr µ10 E×B, derive expressions for He.m. and Se.m. in terms
R
of amplitudes ak,α .
(b) Now quantize the system by replacing the amplitudes via creation and annihilation
√
√ †
operators as follows: ak,α → h
¯a
ˆk,α and a∗k,α → h
¯a
ˆk,α . As a result, the vector potential
becomes operator as well. Use it, as well as commutation relation [ˆ
ak,α , a
ˆ†k0 ,α0 ] = δk,k0 δα,α0 ,
ˆ e.m. and Sˆe.m. .
to derive quantum expressions for operators H
(c) Using E = − ∂A
(remember, we work in the Coulomb gauge), calculate the equal-time
∂t
ˆ t) and E(r
ˆ 0 , t): [A(r,
ˆ t), E(r
ˆ 0 , t)] =?
commutator of operators A(r,
Problem 2. Propagator of the particle moving in the linear potential [15 points]
Particle of mass m moves in one dimension under the influence of constant gravitational
potential mgx. Its Lagrangian is thus L = 21 mx˙ 2 − mgx.
(a) Solve for the classical trajectory xcl (t), which satisfies
d ∂L
dt ∂ x˙ cl
=
∂L
.
∂xcl
Solve this equation
for a path with the boundary conditions xcl (t1 ) = x1 , xcl (t2 ) = x2 , and then calculate the
classical action S[xcl ] =
mg
(t2
2
− t1 )(x2 + x1 ) −
R t2
t1
mg 2
24
dt0 L(xcl (t0 ), x˙ cl (t)) for this path. Show that S[xcl ] =
m(x2 −x1 )2
2(t2 −t1 )
−
(t2 − t1 )3 .
To calculate the path integral, expand path x(t) about the classical one, x(t) =
xcl (t) + δx(t). Demonstrate that D(x2 , t2 ; x1 , t1 ) = J(t2 − t1 )eiS[xcl ] , where J(t2 − t1 ) =
R
D[δx(t)] exp[i
R t2
t1
dt0 m2 (δ x(t
˙ 0 ))2 ] is the free propagator, which was previously found to be
J(t2 − t1 ) = D(0) (0, t2 ; 0, t1 ) =
(b)
mal
Alternative
time
limn→∞
m
2πi
slices
n/2
q
derivation
and
m
.
2πi(t2 −t1 )
consists
performing
the
in
successive
Pn−1 (xj −xj1 )2
Πi=1 −∞ dxi exp i[ m2 j=1
1/2
n−1 R ∞
integrations the result reads
3
m
2πi(k+1)
breaking
−
the
path
integration
Pn−1
j=1
mgxj ].
into
infinitesi-
D(x2 , t2 ; x1 , t1 )
=
Show that after k
im
exp[ 2(k+1)
(xk+1 − x0 )2 − img k2 (xk+1 + x0 ) −
i 24
k(k + 1)(k + 2)mg 2 ], and that in the limit k → ∞ this expression reduces to the result
found in part (a).

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